Chapter 19, Problem 52P

(a)

To determine

The magnetic force on each side of the loop.

Answer to Problem 52P

The magnetic force on each side of the loop is given in table 1.

Explanation of Solution

Refer figure 1.

Write an expression for the force on the side.

  $F=ILB\sin \theta$                                                                                                           (I)

Here,$F$ is the magnitude of the force, $I$ is the magnitude of the current, $L$ is the length of the wire, $B$  is the magnitude of magnetic field and $\theta$  is the angle.

Conclusion:

Consider the top side of the loop:

Substitute $1.0\text{A}$ for $I$, $30.0\text{cm}$ for $L$, $2.5\text{T}$  for $B$ and $60.0°$ for $\theta$ in equation (I) to find $F$.

    $\begin{array}{c}F=\left(1.0\text{A}\right)\left(\left(30.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\left(2.5\text{T}\right)\sin \left(60.0°\right)\\ =\left(1.0\text{A}\right)\left(0.300\text{m}\right)\left(2.5\text{T}\right)\sin \left(60.0°\right)\\ =0.65\text{N}\end{array}$

Consider the right side of the loop:

Substitute $1.0\text{A}$ for $I$, $20.0\text{cm}$ for $L$, $2.5\text{T}$  for $B$ and $30.0°$ for $\theta$ in equation (I) to find $F$.

    $\begin{array}{c}F=\left(1.0\text{A}\right)\left(\left(20.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\left(2.5\text{T}\right)\sin \left(30.0°\right)\\ =\left(1.0\text{A}\right)\left(0.200\text{m}\right)\left(2.5\text{T}\right)\sin \left(30.0°\right)\\ =0.25\text{N}\end{array}$

Consider the bottom side of the loop:

Substitute $1.0\text{A}$ for $I$, $30.0\text{cm}$ for $L$, $2.5\text{T}$  for $B$ and $120°$ for $\theta$ in equation (I) to find $F$.

  $\begin{array}{c}F=\left(1.0\text{A}\right)\left(\left(30.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\left(2.5\text{T}\right)\sin \left(120°\right)\\ =\left(1.0\text{A}\right)\left(0.300\text{m}\right)\left(2.5\text{T}\right)\sin \left(120°\right)\\ =0.65\text{N}\end{array}$

Consider the left side of the loop:

Substitute $1.0\text{A}$ for $I$, $20.0\text{cm}$ for $L$, $2.5\text{T}$  for $B$ and $150°$ for $\theta$ in equation (I) to find $F$.

  $\begin{array}{c}F=\left(1.0\text{A}\right)\left(\left(20.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)\left(2.5\text{T}\right)\sin \left(150°\right)\\ =\left(1.0\text{A}\right)\left(0.200\text{m}\right)\left(2.5\text{T}\right)\sin \left(150°\right)\\ =0.25\text{N}\end{array}$

According to right hand rule, if the thumb, index finger and middle finger are kept perpendicular to each other, the thumb points towards the direction of force if the index finger is kept along the magnetic field and the middle finger is kept along the direction of current.

Thus, the magnetic force on each side of the loop is given in table 1.

Side	$\overrightarrow{L}$	$\theta$	$F\left(\text{N}\right)$	Direction of $\overrightarrow{F}$
Top	$0.300\text{m}\text{right}$	$60.0°$	$0.65$	Into the page
Right	$0.200\text{m}\text{down}$	$30.0°$	$0.25$	Out of the page
Bottom	$0.300\text{m}\text{right}$	$120°$	$0.65$	Out of the page
Left	$0.200\text{m up}$	$150°$	$0.25$	Into the page

Table 1

(b)

To determine

The net magnetic force on the loop.

Answer to Problem 52P

The net magnetic force on the loop is $0$.

Explanation of Solution

Write an expression for the net magnetic force.

  $\Sigma \overrightarrow{F}={\overrightarrow{F}}_{\text{top}}+{\overrightarrow{F}}_{\text{bottom}}+{\overrightarrow{F}}_{\text{right}}+{\overrightarrow{F}}_{\text{left}}$                                                                      (II)

$\Sigma \overrightarrow{F}$ is the net magnetic force, ${\overrightarrow{F}}_{\text{top}}$ is the magnetic force on top side, ${\overrightarrow{F}}_{\text{bottom}}$ is the magnetic force on bottom side, ${\overrightarrow{F}}_{\text{left}}$ is the magnetic force on left side and ${\overrightarrow{F}}_{\text{right}}$ is the magnetic force on right side.

Conclusion:

Substitute $0.65\text{N}\text{in}$ for ${\overrightarrow{F}}_{\text{top}}$, $0.65\text{N}\text{out}$ for ${\overrightarrow{F}}_{\text{bottom}}$, $0.25\text{N}\text{in}$ for ${\overrightarrow{F}}_{\text{left}}$ and $0.25\text{N}\text{out}$ for ${\overrightarrow{F}}_{\text{right}}$

    $\begin{array}{c}\Sigma \overrightarrow{F}=0.65\text{N}\text{in}+0.65\text{N}\text{out}+0.25\text{N}\text{in}+0.25\text{N}\text{out}\\ =0\end{array}$

Thus, the net magnetic force on the loop is $0$.